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walters
Question based on (Groups)
jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings
Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.
am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group
It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?
i am failing to show the axioms
First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.
why do u chose to use |dw:1360875075213:dw|
"\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.
can u also use * if u wnt
For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{-1}\) and \(d=a^{-1}(-b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{-1}ax+ba^{-1}-a^{-1}b=x,\]so we do indeed have inverses.
If you make it clear your group function is composition of functions, you can use * as well.
As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.
let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.
k it is a group but not an abelian group
\[\tau_{10}(x)=1x+0=x\]